How Is Electric Flux Calculated Through a Non-Uniform Field in a Cube?

[Bryon]

Homework Statement

A cube, placed in a region of electric field, is oriented such that one of its corners is at the origin and its edges are parallel to the x, y, and z axes as shown to the left. The length of each of its edges is 1.1 m.

(a) What is the flux through the cube if the electric field is given by E = 3.00y?
3*1.1^3 = 3.993

flux = 3.993Nm^2/C

(b) What is the flux through the cube if the electric field is given by E = -4i + (6 +3y)j
flux = ?

Homework Equations

flux = E dot A = |E||A|cos(theta)

The Attempt at a Solution

-4(1.1)^3 + (6+3*1.1)*1.1^2 = 5.929Nm^2/C

which isn't correct. I though that you have to find the flux in the x and y directions then just sum them up to get the total flux. Do I need to find the angle between the magnitude of the vectors?

 

[gneill]

In (a), the electric field is strictly in the x-direction (parallel to the xy plane), although its magnitude has a y dependency. The important thing, though, is that the direction of the flux lines will be perpendicular to the front and back faces of the cube (where the front and back are assumed to be the faces parallel to the yz plane), and parallel to all other faces.

Things in (b) are a tad more complicated, with the field having distinct x and y (i and j) components. Write the outward pointing unit vectors for the dA area elements for each cube face and take the dot product with the field vector (Cartesian dot product). That will tell you what integrals you'll need to perform to sum up the flux.

 

[Bryon]

Would the dot product be like the following:

[-4i + (6 +3y)j] dot y^2

which would just equal (6+3y)(y^2) then just integrate this from 0 to 1.1?

 

[gneill]

For the front face it would be [-4i + (6 + 3y)j + 0k] dot [1i + 0j +0k] = -4

For the left side face it would be [-4i + (6 + 3y)j + 0k] dot [0i -1j +0k] = -6 - 3y, where y = 0 for the left side face.

Do the rest accordingly.

 

[rwisz]

The flux through the cube would be the same as the flux through any surface enclosing the cube. In this case, the cube is parallel to the x, y, and z axes, so the normal vector of any face would be parallel to the electric field. Therefore, the angle between the electric field and the surface would be 0 degrees, and the cosine of 0 degrees is 1. So the flux through the cube would simply be the magnitude of the electric field multiplied by the area of one face of the cube, which is (1.1 m)^2 = 1.21 m^2. Therefore, the flux through the cube would be:

flux = |E| * A = (|-4i + (6+3y)j|) * 1.21 m^2 = (√(4^2 + (6+3y)^2)) * 1.21 m^2 = (4 + √(36+36y+9y^2)) * 1.21 m^2

Note that in this case, the electric field is not constant throughout the cube, so you cannot simply multiply the magnitude of the electric field by the volume of the cube to get the flux. Instead, you must consider the electric field at each point on the surface of the cube and then integrate over the entire surface to find the total flux.